Lateral Torsional Buckling [Theory and Calculation]

The lateral torsional buckling is the deformation of the beam due to the applied loads away from its longitudinal axis. Further, it causes steel beams failures.

The deformation could occur as translational and rotational movement of the section, and these types of movements are identified as lateral torsional buckling. Figure 1 indicates the deformations that can be seen as a result of lateral torsional buckling.

Figure 01 Lateral Torsional Buckling

As indicated in Figure 01, the beam can be decoded with the application of loads. This deformation could occur laterally and vertically with a rotation of the member. In the design of steel beams, lateral restains are provided in spacings calculated to avoid failures.

Lateral torsional buckling occurs with the increase of loads based on section properties and its restrains. Loading on a beam can not be avoided as it is the purpose of having the beam.

However, the section properties and the restrain conditions can be controlled during construction and design.

• As discussed above too, lateral-torsional buckling occurs when the beam is not full restraints in the lateral direction along the compression flange of the beam.
• The beam is considered full restraint laterally when the connection between the beam and the floor can resist at least a lateral force of 2.5% of the maximum force in the compression flange of the beam. 

When there are no restraints, it is required to provide sections having higher section modulus, if the restraints are provided properly, the size of the beam can be reduced.

It is a proven fact that failures cause the moment of the section away from its axis. Therefore, providing the restrains will definitely reduce the section dimensions. 

However, based on the structural arrangement of the structure, there may not be able to provide the lateral restrains at ends or internally. In such situations, beams need to be designed without considering the lateral restrains.

Mainly, the failures to restrain the compression flange cause the lateral movement of the section. Therefore, lateral-torsional buckling can be avoided by providing internal restains.

Intermediate restraints are provided to reduce the unsupported length in the lateral direction. They should be capable of resisting lateral forces and should have the capacity to retain without deforming. The axial capacity of intermediate restrains shall be checked as per the guidelines in the BS 5950.

Design for Lateral Torsional Buckling of Beam

Section to satisfy the bending requirements, it shall have the bending capacity in the direction of bending (Mc) grater than the applied bending moment and lateral torsional buckling capacity greater the moment generated due to the buckling.

M_x < M_b/m_LT and M_x ≤ M_c

In this article, the procedure needs to be followed in calculating the lateral-torsional buckling capacity is discussed. And the article steel beam design to bs 5950 could be referred bending capacity checks. 

Lateral Torsional Buckling resistance ( M_b/m_LT ) can be calculated as illustrated below. Two methods used to evaluate the Buckling Resisting Moment ( M_b ). Based on the preference of the designer, either method could be used.

1. Rigorous method
2. Simplified method

When both methods are compared, it seems that the main difference is the method of evaluating the Bending Strength ( P_b ).

A detailed explanation of the method of designing as a steel beam is discussed in the article steel beam design worked example. 

Check Lateral Torsional Buckling

Lateral torsional buckling Example

Data:

• Consider simply supported beam with not intermediate restraints
• Beam span 6m
• Maximum design bending moment 100 kNm

As discussed above there are two methods for checking the lateral-torsional buckling. Let’s discussed them with a worked example.

It is required to satisfy the following equation for a section to be ok for buckling.

M_x < M_b / m_LT

For simplicity, In this example, no intermediate restraints are considered.

Then, 

m_LT = 0.925, Table 18, BS 5950

M_b = P_b S_x Cl. 4.3.6.4

Firstly let’s check the buckling capacity with rigorous method.

The following section data are considered in the calculation

• D = 500 mm
• T = 16 mm
• t = 10 mm
• B = 200 mm
• b = 100 mm
• r₁ = 20 mm
• d = 500 – 16 x 2 – 2 x 20 = 428 mm
• S_x = 2175×10³ mm³
• Z_x = 1914×10³ mm³
• r_y = 43.3 mm
• Section is plastic as per its dimensions

Rigorous Method

M_b = P_b S_x

P_b is fuction of  λ_LT and P_y

λ_LT = uvλ√(β_w)

λ = L_E / r_y

L_E – can be found from Table 13 (Cl. 4.3.5.1) and consider L_LT = L – span

Thus,

L_E = 1.0 L_LT = 1 x 6 = 6 m

λ = L_E / r_y = 6000 / 43.3 = 138.568

For rolled I and H sections, Cl. 4.3.6.8

x = D / T used with u = 0.9

x = D / T = 500 / 16 = 31.25

β_w can be be obtained from the Cl 4.3.6.9

β_w = 1 for Class 1 plastic or Class 2 Compact sections

v – slenderness factor – obtained from Table 19 as per the λ / x and η

λ / x = 138.568 / 31.25 = 4.434

For equal flanges η = 0.5

v = 0.84 from Table 19 [conservative value taken; λ / x  = 4.5, interpolation shall be used for exact value]

λ_LT = uvλ√(β_w) = 0.9 x 0.84 x 138.568 x √(1) = 104.8

λ_LO to be obtained from Table 16 (indicated at the bottom of the Table 16)

If λ_LO ≥ λ_LT ; P_b = P_y or Othewise P_b shall be taken from Table 16 for rolled sections.

If λ_LO ≥ λ_LT no allowance needs to be made for lateral-torsional buckling and otherwise check for lateral-torsional buckling.

P_y = 275 N/mm² ; λ_LO = 34.3

λ_LO < λ_LT

Hence, check for lateral torsional buckling

From Table 16, for λ_LT = 104.8 ; P_b = 117 N/mm²

M_b = P_b S_x = 117 x 2175 x 10³ x 10^-6 = 254.5 kNm

M_b / m_LT = 254.8 / 0.925 = 275.4 kNm

Therefore, M_x = 100 kNm < M_b / m_LT = 239.838 kNm

Section is adquate for lateral torsional buckling as per the rigorous method.

Simplified Method

We need not doing both the calculation to check the buckling resistance. 

M_b = P_b S_x : Cl. 4.3.7

The determination of the P_b is not the same as the rigorous method.

This method provides conservative answeres.

P_b can be obtained from Table 20 of BS 5950 as per √(β_w) (L_E / r_y ) and D / T

β_w = 1 ; as previous calculation.

L_E / r_y = 138.568 ; from above calculations

√(β_w) (L_E / r_y ) = 1^0.5 x 135.568 = 138.568

D / T = 500 / 16 = 31.25

Now, can obtain from Tabe 20

P_b = 116.646 N/mm²  

M_b = P_b S_x = 116.646 x 2175 x 10³ x 10^-6 = 253.705 kNm

M_b / m_LT = 253.705 / 0.925 = 274.3 kNm

Thus, M_x = 100 kNm < M_b / m_LT = 274.3 kNm

Therfore, the section is adequate.